Question 1 The reaction: X (g) + e- X- (g) + energy Question 2 4.000 moles of Cl
ID: 592607 • Letter: Q
Question
Question 1
The reaction: X (g) + e- X- (g) + energy
Question 2
4.000 moles of Cl2 (g) react with excess P4 (s) to produce 0.5741 mol of phosphorus trichloride. What is the percent yield for this reaction?
P4 (s) + 6 Cl2 (g) 4 PCl3 (l)
Question 3
If 239 grams of O2 are produced in the reaction below, how many grams of HgO have reacted?
2 HgO (s) 2 Hg (l) + O2 (g)
Question 4
8.84 g of a compound is analyzed and found to contain 3.53 g of C, 0.295 g of H, 4.08 g of N, and 0.935 g of O. Which of the following could be the molecular formula for this compound?
represents the first ionization energyExplanation / Answer
Question 1
The reaction: X (g) + e- X- (g) + energy
represents electron affinity >>>>>>answer
Question 2
P4 (s) + 6 Cl2 (g) 4 PCl3 (l)
6 moles of Cl2 react with P4 to gives 4 moles of PCl3
4 moles of Cl2 react with P4 to gives = 4*4/6 = 2.67 moles of PCl3
Theoritical yiled = 2.67moles of PCl3
percentage yield = actual yield*100/theoritical yield
= 0.5741*100/2.67 = 21.53% >>>>answer
question3
2 HgO (s) 2 Hg (l) + O2 (g)
1 mole of O2 produced from 2 moles of HgO
32g of O2 produced from 2*216.58 g of HgO
239 g of O2 produced from = 2*216.58*239/32 = 3240g of HgO >>>answer
question 4
Element wt At.wt no of moles simple ratio
C 3.53 g 12 3.53/12 = 0.294 0.294/0.0584 = 5
H 0.295g 1 0.295/1 = 0.295 0.295/0.0584 = 5
N 4.08g 14 4.08/14 = 0.291 0.291/0.0584 = 5
O 0.935 16 0.935/16 = 0.0584 0.0584/0.0584 = 1
empirical formula = C5H5N5O
C10H10N10O2 >>>>>answer
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